How do I evaluate $\sum_{r=1}^{n} [r(r+1)(r+2)(r+3)] $?

Question

I need to evaluate: \begin{equation} \sum_{r=1}^{n} [r(r+1)(r+2)(r+3)] \end{equation}

But I don't know where to start. Can someone possibly give a hint?

I need to solve this using telescopic series.

Compute answers for small $n$. Then try to guess the general answer. Then try to prove it. — Grigory M, Jan 03 '15 at 14:31
You could also expand it in terms of $r^4$, $r^3$, $r^2$, $r$ and $r^0$ and then use the formulas for those sums separately. — Slugger, Jan 03 '15 at 14:33
@Slugger: Can I do it without using the sum formula for $r^4$ — user63762453, Jan 03 '15 at 14:35
Do you know how to telescope $r(r+1)(r+2)(r+3)$ and $r(r+1)(r+2)$? — hrkrshnn, Jan 03 '15 at 14:41
@HatMan Sorry, a typo, the question was: $r(r+1)(r+2)(r+3)$ instead of $r(r+1)(r+2)(r+2)$ — user63762453, Jan 03 '15 at 14:44

Jack D'Aurizio · Answer 1 · 2015-01-03T15:28:33.450

9

$r(r+1)(r+2)(r+3) = 24\binom{r+3}{4}$, and since $$ \sum_{k=j}^{N}\binom{k}{j}=\binom{N+1}{j+1}$$ holds by induction, we have: $$ \sum_{r=1}^{n} r(r+1)(r+2)(r+3) = 24\sum_{r=1}^{n}\binom{r+3}{4} = 24\binom{n+4}{5} = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$

edited Jan 03 '15 at 15:28

answered Jan 03 '15 at 14:51

Jack D'Aurizio

353,855

2

... and this formula may be how David found his mysterious answer. – GEdgar Jan 03 '15 at 15:30

score 7 · Accepted Answer · answered Jan 03 '15 at 14:45

7

$$ r(r+1)(r+2)(r+3) = \frac15\left((r+1)(r+2)(r+3)(r+4)(r+5)- r(r+1)(r+2)(r+3)(r+4)\right) \\ $$

answered Jan 03 '15 at 14:45

David Holden

18,040

Sorry I had a typo, can you please see the updated question? – user63762453 Jan 03 '15 at 14:48
thx, have adjusted the suggestion – David Holden Jan 03 '15 at 14:49
Okay, got it thank you.. – user63762453 Jan 03 '15 at 14:50

GPerez · Answer 3 · 2015-01-03T15:02:36.470

I suggest expanding the product to get $$\sum_{r=1}^n(r^4+6r^3+11r^2+6r) = \\ =\sum_{r=1}^n r^4 + 6\sum_{r=1}^n r^3 + 11\sum_{r=1}^n r^2 + 6\sum_{r=1}^n r$$ Now the problem is finding formulas for these sums. The linear sum you probably know (or should) and you might know the sum of squares too. For the ones you don't, it's useful to know that the sum $\sum_{r=1}^n r^k$ is a polynomial in $n$ of degree $k+1$. Therefore if you calculate the first $k+1$ sums, you'll get a linear system of equations in the coefficients (to be determined) of the polynomial.

For example, take $S(n) = \sum_{r=1}^nr^2$. We conjecture that this is equal to $an^3+bn^2 + cn + d$. First of all note that $d = 0$, necessarily. Next: $$\begin{align}S(1) &= 1 &= & a + b + c \\ S(2) &= 5 &= & 8a + 4b + 2c \\ S(3) &= 14 &= &27a + 9b + 3c\end{align}$$ Use your preferred method to solve the system. The same will go for the other sums.

This is quite a brute force method and as such it'll be tedious, but it's good to have in your arsenal of strategies. Additionally, you can skip the intermediate step of calculating the first $n$ squares, cubes, etc. by conjecturing from the very beginning that the sum is a polynomial of degree 5.

Edit: I now see you need to use the telescoping sum, so disregard if you wish. I will leave the answer anyway for any who might find it useful.

Thanks for answering :D Atleast i got do know how to derive sum formulas. I already marked it +10 — user63762453, Jan 06 '15 at 16:20

How do I evaluate $\sum_{r=1}^{n} [r(r+1)(r+2)(r+3)] $?

3 Answers3